What is the angle between <-4,3,-8 > and < 1,-1,5>?

Oct 8, 2016

theta ~~ 2.56 radians or 146.76°

Explanation:

The dot product is used, because it has two definitions:

barA•barB = (A_1)(B_1) + (A_2)(B_2) + (A_3)(B_3)

and

barA•barB = |barA||barB|cos(theta)

where $\theta$ is the angle between the two vectors.

Compute the dot product, using the first definition:

barA•barB = (-4)(1) + (3)(-1) + (-8)(5) = -41

Compute the magnitude of vector A:

|barA| = sqrt((-4)² + 3² + (-8)²) = sqrt(89)

Compute the magnitude of vector B:

|barB| = sqrt(1² + (-1)² + 5²) = sqrt(27)

Substitute the above into the second equation:

$- 41 = \left(\sqrt{89}\right) \left(\sqrt{27}\right) \cos \left(\theta\right)$

Solve for $\theta$:

$\theta = {\cos}^{-} 1 \left(- \frac{41}{\sqrt{89} \sqrt{27}}\right)$

theta ~~ 2.56 radians or 146.76°